11А. Найдите все значения параметра a, при каждом из которых уравнение \({x^2}-a\,x + 2 = 0\) имеет два различных корня, принадлежащих интервалу \(\left( {0;\,3} \right)\).
\(\left\{ {\begin{array}{*{20}{c}}{D > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{0 < {x_B} = -\frac{b}{{2a}} < 3,}\\{f\left( 0 \right) > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 3 \right) > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{D = {a^2}-8 > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{0 < \frac{a}{2} < 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 0 \right) = 2 > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 3 \right) = 9-3a + 2 > 0}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a\, \in \,\left( {-\infty ;-\sqrt 8 } \right) \cup \left( {\sqrt 8 ;\infty } \right),}\\{a\, \in \,\left( {0;6} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;\frac{{11}}{3}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,a\, \in \,\left( {2\sqrt 2 ;\frac{{11}}{3}} \right).\,\,\) ОТВЕТ: \(\left( {2\sqrt 2 ;\,\frac{{11}}{3}} \right).\)
Введём функцию \(f\left( x \right) = {x^2}-ax + 2\) графиком которой является парабола ветвями вверх. Для того чтобы уравнение имело два различных корня, принадлежащих интервалу \(\left( {0;3} \right)\), необходимо выполнение следующих условий (см. рис.):